3y^2-12y-16=0

Solution for 3y^2-12y-16=0 equation:

3y^2-12y-16=0

a = 3; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·3·(-16)
Δ = 336
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{21}}{2*3}=\frac{12-4\sqrt{21}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{21}}{2*3}=\frac{12+4\sqrt{21}}{6}
$